scala中的compose和andThen函数剖析

源码

/** Composes two instances of Function1 in a new Function1, with this function applied last.   *   *  @tparam   A   the type to which function `g` can be applied   *  @param    g   a function A => T1   *  @return       a new function `f` such that `f(x) == apply(g(x))`   */  @annotation.unspecialized def compose[A](g: A => T1): A => R = { x => apply(g(x)) }  /** Composes two instances of Function1 in a new Function1, with this function applied first.   *   *  @tparam   A   the result type of function `g`   *  @param    g   a function R => A   *  @return       a new function `f` such that `f(x) == g(apply(x))`   */  @annotation.unspecialized def andThen[A](g: R => A): T1 => A = { x => g(apply(x)) }

源码中compose是说实现一个新的函数f(x) == apply(g(x))的函数，本质就是
构造一个新的函数为先运行f再运行g
而andThen则是实现新的函数f(x) ==g(apply(x))，二这个本质就是先运行g再运行f

/**  * 传入Int,结果翻倍  */val doubleNum :Int => Int  = {num =>  val numDouble = num * 2  println(s"double Num ($num * 2 = $numDouble)")  numDouble}/**  *传入Int,结果加一  */val addOne:Int => Int = {num=>  val sumNum = num + 1  println(s"add One ($num + 1 = $sumNum)")  sumNum}

/**  *  * @param g  * @param f  * @tparam A  * @tparam B  * @tparam C  * @return  */def myAndThen[A, B, C](g: A => B, f: B => C): A => C = x => f(g(x))/**  *  * @param g  * @param f  * @tparam A  * @tparam B  * @tparam C  * @return  */def myCompose[A, B, C](g: A => B, f: C => A): C => B = x => g(f(x))

compose：
期待一个C类型转化为B类型，也就是是说compose则是将先执行靠里面的函数，再将结果传给外层函数，所以先执行f将C类型的x转化为一个A类型，再作为参数传给g，得到一个B类型。
andThen：
期待一个A类型转化成一个C类型，x是一个A类型，将x传给g，作为g的参数得到一个一个B类型，再将结果作为参数传给f，f在是将B类型转化为一个C类型，最终实现A类型的x转化为C类型。
测试用例：

val myComposeTest = myCompose[Int, Int, Int](addOne, doubleNum)myComposeTest(2)val composeDoubleNumInAddOne = addOne compose doubleNumval composeAddOneInDoubleNum = doubleNum compose addOnecomposeDoubleNumInAddOne(2)composeAddOneInDoubleNum(2)assert(myComposeTest(2)==composeDoubleNumInAddOne(2))val myAndThenTest = myAndThen[Int,Int,Int](addOne,doubleNum)myAndThenTest(2)val addOneAndThenDoubleNum = addOne andThen doubleNumval doubleNumAndThenAddOne = doubleNum andThen addOneaddOneAndThenDoubleNum(2)doubleNumAndThenAddOne(2)assert(myAndThenTest(2)==addOneAndThenDoubleNum(2))

效果